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\(A=\frac{10^{2015}-1}{10^{2016}^{ }-1}=\frac{10^{2015}}{10^{2016}}=\frac{1}{1},B=\frac{10^{2014}-1}{10^{2015}-1}=\frac{10^{2014}}{10^{2015}}=\frac{1}{1}A=B\Rightarrow\)
b, 2000A = \(\frac{2000\left(2000^{2015}+1\right)}{2000^{2016}+1}\)
= \(\frac{2000^{2016}+2000}{2000^{2016}+1}\)
= \(\frac{\left(2000^{2016}+1\right)+1999}{2000^{2016}+1}\)
= \(\frac{2000^{2016}+1}{2000^{2016}+1}\) + \(\frac{1999}{2000^{2016}+1}\)
= 1 + \(\frac{1999}{2000^{2016}+1}\)
2000B = \(\frac{2000\left(2000^{2014}+1\right)}{2000^{2015}+1}\)
= \(\frac{2000^{2015}+2000}{2000^{2015}+1}\)
= \(\frac{\left(2000^{2015}+1\right)+1999}{2000^{2015}+1}\)
= \(\frac{2000^{2015}+1}{2000^{2015}+1}\) + \(\frac{1999}{2000^{2015}+1}\)
= 1 + \(\frac{1999}{2000^{2015}+1}\)
So sanh
câu b tiếp
So sánh 2000A với 2000B
Vì \(\frac{1999}{2000^{2016}+1}\) < \(\frac{1999}{2000^{2015}+1}\)
→ 2000A< 2000B
→ A<B
10A=(10^2014+1).10/10^2015+1=10^2015+10/10^2015+1=10^2015+1+9/10^2015+1=1+(9/10^2015+1) 10B=(10^2015+1).10/10^2016+1=10^2016+10/10^2016+1=10^2016+1+9/10^2016+1=1+(9/10^2016+1) Vì 9/10^2015+1>9/10^2016+1 nên 10A>10B .Từ đó suy ra A>B
xét A ta có
\(10A=\frac{10.\left(10^{2014}+1\right)}{10^{2015}+1}=\frac{10^{2015}+10}{10^{2015}+1}=\frac{\left(10^{2015}+1\right)+9}{10^{2015}+1}\)suy ra \(10A=1+\frac{9}{10^{2015}+1}\)
xét B ta có
\(10B=\frac{10.\left(10^{2015}+1\right)}{10^{2016}+1}=\frac{10^{2016}+10}{10^{2016}+1}=\frac{\left(10^{2016}+1\right)+9}{10^{2016}+1}=1+\frac{9}{10^{2016}+1}\)
Vì 10A>10B suy ra A >B
10A = 10 2015 + 1 10. 10 2014 + 1
= 10 2015 + 1 10 2015 + 10
= 10 2015 + 1 10 2015 + 1 + 9
suy ra 10A = 1 + 10 2015 + 1 9
Vì \(\frac{10^{2014}+1}{10^{2015}+1}< 1\Rightarrow B=\frac{10^{2014}+1}{10^{2015}+1}< \frac{10^{2014}+1+9}{10^{2015}+1+9}\)
\(\Rightarrow B< \frac{10^{2014}+10}{10^{2015}+10}\)
\(\Rightarrow B< \frac{10\left(10^{2013}+1\right)}{10\left(10^{2014}+1\right)}\)
\(\Rightarrow B< \frac{10^{2013}+1}{10^{2014}+1}\)
\(\Rightarrow B< A\)
Vậy A > B
A=10^2014+1/10^2015+1
10A=10^2015+10/10^2015+1
10A=10^2015+1+9/10^2015+1
10A=1+(9/10^2015+1)(1)
B làm tương tự (2)
Từ (1); (2)
Suy ra 10A>10B
Suy ra A>B
Vậy........
Vi B < 1 nen ta co :
\(B=\frac{10^{2015}+1}{10^{2016}+1}< \frac{10^{2015}+1+9}{10^{2016}+1+9}\)
\(\Rightarrow B< \frac{10^{2015}+10}{10^{2016}+10}=\frac{10\left(10^{2014}+1\right)}{10\left(10^{2015}+1\right)}=A\)
Vay \(B< A\)
có :
\(B=\frac{10^{2015}+1}{10^{2014}+1}>1\)
\(\Rightarrow\frac{10^{2015}+1}{10^{2014}+1}>\frac{10^{2015}+1+9}{10^{2014}+1+9}\) \(=\frac{10^{2015}+10}{10^{2014}+10}=\frac{10.\left(10^{2014}+1\right)}{10.\left(10^{2013}+1\right)}\)
\(=\frac{10^{2014}+1}{10^{2013}+1}=A\)
\(\Rightarrow B>A\)
Vậy B > A
k cho mk nhé
Bài 1 :
Ta có :
\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)
Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)
Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)
Vậy \(A>B\)
Bài 2 :
Ta có :
\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)
\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)
\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)
\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)
Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)
Nên : \(M>4\)
Vậy \(M>4\)
Bài 3 :
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)
Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)
\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)
\(\Rightarrow A< \frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
Bài 4 :
\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)
\(\Rightarrow A=\frac{1008}{2017}\)
Vậy \(A=\frac{1008}{2017}\)
\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)
\(1-\frac{1}{x+2}=\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)
\(\Rightarrow x+2=2017\)
\(\Rightarrow x=2017-2=2015\)
Vậy \(x=2015\)
Ta có công thức :
\(\frac{a}{b}>\frac{a+c}{b+c}\)\(\left(\frac{a}{b}>1;a,b,c\inℕ^∗\right)\)
Áp dụng vào ta có :
\(B=\frac{10^{2016}+1}{10^{2015}+1}>\frac{10^{2016}+1+9}{10^{2015}+1+9}=\frac{10^{2016}+10}{10^{2015}+10}=\frac{10\left(10^{2015}+1\right)}{10\left(10^{2014}+1\right)}=\frac{10^{2015}+1}{10^{2014}+1}=A\)
\(\Rightarrow\)\(B>A\) hay \(A< B\)
Vậy \(A< B\)
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